So, taking into account the bounce loss as 'k', and rotor mass
as 'M3' the formula would be:
k*M1*g*h = M1*g*d + M2*g*d + M3*.25*r^2*w^2
solving for 'd' = (k*M1*g*h - M3*.25*r^2*w^2)/(M1 + M2)*g
notice in general the (k*M1-M3)/(M1+M2) relation. Let's just
substitiue for 1kg and 5 kg weights.... (k-x)/(6) also let's
look at 2kg he used. (2k-x)/(7). Notice that if x were close
to 1kg things look bleak, also if k were .5 things look
bleak. Also, when you select a heavier M1 that may match
the load better, the k factor will become closer to 1, just
like VSWR matching the load impedance.
So in short, I don't believe he did his experiment correctly.
First he should have found out why the drop only lifted the
heavy mass 4cm - instead of his 20cm he thought was right.
v/r Ken Carrigan
Physical laws can not be ignored...